2LB_Lab_4_EfficiencyBulb4

2LB_Lab_4_EfficiencyBulb4 - Efficiency of a Light Bulb (Pre...

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Efficiency of a Light Bulb (Pre Lab Assignment Included) Objective: An ordinary incandescent light bulb converts electrical energy into both light and heat. In the light bulb, current is made to flow through a very thin tungsten wire, causing it to get extremely hot and glow white. The wire is kept in a glass bulb which has been evacuated to prevent air from chemically reacting with it. The light energy is radiated away through the clear glass envelope, while the heat energy raises the temperature of the bulb. This week we will measure both the light and heat energy output of a light bulb. From our measurements we will calculate what fraction of the total energy goes into light – the "efficiency" of the light bulb. Note that this definition of efficiency is based on our interest, in this case, the amount of light output. Apparatus 1. Computer with Data Studio software 2. PASPort temperature sensor 3. Light bulb assembly with electrical power supply 4. Digital voltmeter/ammeter 5. Support stand to hold light bulb and thermometer 6. Clear plastic vial to serve as a calorimeter 7. Electric-powered magnetic stirrer 8. Magnetic stirring rod ( caution: do not lose this in the sink) 9. Hot water and room temperature water 10. A 250 ml plastic beaker to fetch and weigh the water 11. Black ink
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Introduction to the Physics The glowing incandescent light bulb gives off heat and light. We can measure the heat given off using a simple calorimeter filled with water. The energy consumption per second of a light bulb is measured in terms of power using the units Watts = Joules/sec . The heat energy given off in time t is Δ Q = P heat t (1) where P heat is the rate at which thermal energy is given off by the light bulb. If the total heat capacity of the calorimeter (water plus container plus light bulb) is C , then Δ Q = C ( T T 0 ) (2) where T 0 is the starting temperature of the system, and T is the temperature at time t . The heat capacity of the total calorimeter will have a contribution from the water of m water c water and a part from the container and light bulb which we will call C cal , so by energy conservation, P heat t Q = ( m water c water + C cal )( T T 0 ) (3) where we have used the equality that the heat energy lost by the bulb is gained by the water and other parts of the calorimeter. Now differentiating Eq. (3) with respect to t we get, dT dt = P heat ( m water c water + C cal ) . (4) Thus if we plot the temperature T as a function of the time t , the slope is given by Eq. (4).
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This note was uploaded on 01/11/2010 for the course PHYS 2b taught by Professor Clare during the Fall '07 term at UC Riverside.

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2LB_Lab_4_EfficiencyBulb4 - Efficiency of a Light Bulb (Pre...

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